Therefore, λ 2 is an eigenvalue of A 2, and x is the corresponding eigenvector. Expert Answer . This ends up being a cubic equation, but just looking at it here we see one of the roots is 2 (because of 2−λ), and the part inside the square brackets is Quadratic, with roots of −1 and 8. Enter your solutions below. or e 1, e 2, … e_{1}, e_{2}, … e 1 , e 2 , …. :2/x2 D:6:4 C:2:2: (1) 6.1. n is the eigenvalue of A of smallest magnitude, then 1/λ n is C s eigenvalue of largest magnitude and the power iteration xnew = A −1xold converges to the vector e n corresponding to the eigenvalue 1/λ n of C = A−1. v; Where v is an n-by-1 non-zero vector and λ is a scalar factor. Qs (11.3.8) then the convergence is determined by the ratio λi −ks λj −ks (11.3.9) The idea is to choose the shift ks at each stage to maximize the rate of convergence. 1. A number λ ∈ R is called an eigenvalue of the matrix A if Av = λv for a nonzero column vector v ∈ … Then the set E(λ) = {0}∪{x : x is an eigenvector corresponding to λ} detQ(A,λ)has degree less than or equal to mnand degQ(A,λ)